Calculus Trig Derivatives ?

F'(x)=2 ( cos^4 2x - x*4*2 sin (2x) cos^3 (2x) )

= 2 cos^4(2x) - 16 x sin (2x) cos^3 (2x).

B) F'(x)=(3x+8)^2(4x-5)^4=

6(3x+8) (4x-5)^4 + 16 (3x+8)^2 (4x-5)^3=

= 2( 9x +24 +24 x +64 ) (3x+8)(4x-5)^3=

2(35x + 88) (3x+8)(4x-5)^3

The first part should be treated like this. F(x) = f(x)g(x). g(x) = h(i(j(x))). Where h(x) = x^4, i(x) = cos(x), and j(x) = 2x. Using the chain rule, g'(x) = h'(i(j(x)) * i'(j(x)) * j'(x). you got the second part right (cos^42x)(2), but the first part should be (-16xcos^3(2x)sin(2x)) which was simplified from (2x)(4cos^3(2x))(-sin(2x))(2).

The second one has the same idea. 6(3x+8)(4x-5)^4 + 16(4x-5)^3(3x+8)^2

you did not finished with the chain rule

f(x) = cos^4 2x

f'(x) = 4 cos^3 2x (-sin 2x) (2)

B. use product rule

2(3x+8)(3) + 4(4x-5)^3(4)

6(3x+8) + 16(4x-5)^3

Ouch. A is a PITA.

Your product rule was right, but you need a double chain rule for the trig part.

u = 2x

F = u(cos^4)u

v = cosu

F = uv^4

So F' =

u'(v^4) + (4(v^3)v'u')u

factor out u'v^3, sub for v

= u'[(cosu)^3][cosu + 4(-sinu)u]

sub for x

F'(x) = 2(cos2x)^3(cos2x - 8xsin2x)

Nothing I can do will make that less ugly.

B is easier, just two single chain rules

u = 3x + 8

v = 4x - 5

u^2v^4 derive =

= 2uu'(v^4) + 4v'(4v^3)u^2

= 2uv^3(vu' + 2uv')

= 2(3x+8)(4x-5)^3[3(4x-5) + 8(3x+8)]

= 2(3x+8)(4x-5)^3(36x+49)

Another ugly one.